SOLUTION: A donut baker wants to ensure that the radius of its donuts is within +- 0.25cm. Based on past statistical studies, she knows that the donut�s radius has a standard deviation of 1.

Question 1083276: A donut baker wants to ensure that the radius of its donuts is within +- 0.25cm. Based on past statistical studies, she knows that the donut�s radius has a standard deviation of 1.25cm. How many measurements should she take to ensure a 90 percent confidence in the radius of the donuts are within 0.25cm, rather than measure each and every donut that is baked?
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
i think it works this way.

at 90% confidence level, your critical z-score will be plus or minus 1.644853626.

i used the ti-84 plus calculator to find this.
90% confidence level using a 2-tailed distribution leaves an alpha of.05 on each end.
the plus z-score gives you an alpha of .05 on the right end of the distribution.
the minus z-score gives you an alpha of.05 on the left end of the distribution.
add up both tails and you get an overall alpha of .10 which give you 90% of the normal distribution within the middle.

the formula for sample z-score is z = (x-m)/se

z is the z-score
x is the raw score
m is the mean
se is the standard error

the standard error is calculated by the formula of se = sd / sqrt(n)

sd is the standard deviation
n is the sample size

if you replace se with sd / sqrt(n), then the formula for z-score becomes z = (x-m) / (sd / sqrt(n))

this is equivalent to z = (x-m) * sqrt(n)) / sd

at the limits, you want (x-m) to be equal to plus or minus .25.

therefore, the formula will become:

1.644853626 = .25 * sqrt(n) / 1.25

that's because the critical z-score is 1.644853626 and the standard deviation is 1.25

solve for sqrt(n) to get sqrt(n) = 1.644853626 / .25 * 1.25 = 8.22426813

solve for n to get n = 67.63

you would round this up to 68.

the sample size should be at least 68 in order to ensure that the limits are within plus or minus .25 at 90% confidence level.

for example:

assume the sample mean is 10 cm.

the lower limit would have to be 9.75 cm and the upper limit would have to be 10.25 cm.

your standard deviation is 1.25

your standard error is 1.25 / sqrt(68) = .1525847656

your lower z-score would be (9.75-10)/.1525847656 = -1.64924225

your upper z-score would be (10.25-10)/.1525847656 = 1.64924225

a z-score of-1.64924225 gives you an alpha of .0495489908 on the left end of the distribution curve.

a z-score of 1.64924225 gives you an alpha of .0495489908 on the right end of the distribution curve.

that's a 90% confidence interval.

90% of your sample means will be between these limits.

if you had rounded the critical z-score to plus or minus 1.645, you would have gotten the same result.

for example:

1.645 = .25/1.25 * sqrt(n)

sqrt(n) = 1.645 * 1.25 / .25 = 8.225

8.225^2 = 67.670625 which rounds up to 68.

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